Question

How to find Solution to differential equation
(x - 4) y^{4} dx - x^{3} (y^{2} - 3 ) dy = 0

Answer 1

How to find Solution to not exact differential equation
\[(x - 4) y^{4} dx - x^{3} (y^{2} - 3 ) dy = 0\]

Answer 2

Isn't \[\int\limits \frac{x-4}{x^3}dx=\int\limits \frac{y^2-3}{y^4}dy\] enough?

Answer 3

It's probably not, I'm still pants at DE

Answer 4

Anyway: don't ask me HOW to integrate that (use Wolfram integrator if you just want the answer), it's probably some counter-intuitive trig substitution.

Answer 5

i got
\[- \frac{ 1 }{ x } + \frac{ 2 }{ x^{2} }-\frac{ 1 }{ y } + \frac{ 1 }{ y^{3} } = C\]
am i right?

Answer 6

http://integrals.wolfram.com/index.jsp?expr=%28x-4%29%2Fx%5E3&random=false
http://integrals.wolfram.com/index.jsp?expr=%28x%5E2-3%29%2Fx%5E4&random=false

So your signs are wrong for either the x or y terms but it's fine other than that.

Answer 7

@gerryliyana
in this type of differntial eqution, we use separation of variable technique
as we can see from this differential equation, variables can be separated easily

Answer 8

@ gerryliyana i m getting
-1/x + 2/x^2 + 1/ y^2 - 1 /y^3 = C
where C is constant

Answer 9

@gerryliyana right????

Answer 10

ok @niksva
but i got
-1/x + 2/x^2 + 1/ y - 1 /y^3 = C
where C is constant

Answer 11

omg my mistake it is 1/y

Answer 12

no problem :)